Solving the Fourier Series of a 2π-Periodic Function

[struggles]

Homework Statement

The odd 2π-periodic function f(x) is defined by
f(x) = x² π > x > 0
-x² −π<x<0
Find the coefficient b_n in the Fourier series
f(x) = a₀/2 + ∑(a_n cos(nx) + b_n sin(nx)).
What are the values of the coefficients a₀ and a_n and why?

Homework Equations

b_n = 1/π ∫ f(x)sin(nx)
a_n = 1/π ∫f(x)cos(nx)

The Attempt at a Solution

[/B]
I'm unsure what to do as the function changes so you cannot integrate between pi and -pi like you would normally. Also as the function is even would b_n = 0?

 

[Fightfish]

I'm unsure what to do as the function changes so you cannot integrate between pi and -pi like you would normally.

The function doesn't "change" - it is defined in a piecewise manner. So, you just have to break up the integration interval accordingly.

Also as the function is even would b_n = 0?

Are you sure the function is even?

 

[Samy_A]

Redudant with @Fightfish 's post.

 

[Ray Vickson]

Homework Statement

The odd 2π-periodic function f(x) is defined by
f(x) = x² π > x > 0
-x² −π<x<0
Find the coefficient b_n in the Fourier series
f(x) = a₀/2 + ∑(a_n cos(nx) + b_n sin(nx)).
What are the values of the coefficients a₀ and a_n and why?

Homework Equations

b_n = 1/π ∫ f(x)sin(nx)
a_n = 1/π ∫f(x)cos(nx)

The Attempt at a Solution

[/B]
I'm unsure what to do as the function changes so you cannot integrate between pi and -pi like you would normally. Also as the function is even would b_n = 0?

What are the integration limits? [You should always indicate this; it is an important part of your analysis.] Have you never before performed integrations of piece-wise funcctions (defined by different formulas in different regions? Why do you say in your final sentence that the function is even?

 

[struggles]

so would i just break it up to get b_n = 1/π[∫^π₀ x²sin(nx) + ∫⁰_-π -x²sin(nx)]

 

[Samy_A]

so would i just break it up to get b_n = 1/π[∫^π₀ x²sin(nx) + ∫⁰_-π -x²sin(nx)]

Yes, and if you give it a little thought, computing one of the integrals (for each n) is all the work you need to do.

 

[struggles]

so when integrating the 2 integrals are the same except for the sign changes. This means when n = odd I get b_n = 2nπ² - 8n³
and when n = even b_n = -2nπ². Does that sound feasible?

 

[Samy_A]

so when integrating the 2 integrals are the same except for the sign changes. This means when n = odd I get b_n = 2nπ² - 8n³
and when n = even b_n = -2nπ². Does that sound feasible?

As your function is odd, $\displaystyle b_n=\frac{2}{\pi}\int_0^{\pi} x²\sin(nx) \, dx$.
I get something different for the $b_n$, but I can make a computation error just as you can.

 

[struggles]

I've re-done it to get (-1)ⁿ(-π²n-2n³)-(2n³) Thank you for all your help. Also would a_n = 0?

 

[Samy_A]

I've re-done it to get (-1)ⁿ(-π²n-2n³)-(2n³) Thank you for all your help

Still different from what I got, but again, you may have the correct result. Difficult to check without seeing the calculation.

Also would a_n = 0?

$a_n$ is the definite integral of an odd function over $(-\pi,+\pi)$. The integrals between $(-\pi,0]$ and $[0,+\pi)$ cancel out. So yes, $a_n=0$.

 

[Samy_A]

I've re-done it to get (-1)ⁿ(-π²n-2n³)-(2n³) Thank you for all your help. Also would a_n = 0?

Come to think of it, the Fourier coefficients of a continuous function (among others, but that is of no importance here) on $[-\pi, +\pi]$ satisfy a number of convergence properties.
For example, $\displaystyle \lim_{n\rightarrow +\infty}{a_n}=0,\ \lim_{n\rightarrow +\infty}{b_n}=0,\ \sum_{n=0}^\infty { |a_n|}^{2}<\infty, \sum_{n=1}^\infty { |b_n|}^{2}<\infty$.

Clearly the $b_n$ you got do not satisfy these properties.